A binary tree with n nodes has exactly n+1 null nodes. So, if number of nodes are 20 then number of null nodes will be 21. So, 21 is correct.  

Follow the following formula to calculate number of nodes.²ⁿC_n / (n+1) where n is the number of nodes.^2*3C₃ / (3+1)=5

Quick sort choose one element and divide the list into two. It uses divide and conquer method. A key element is chosen then it is compared and the element is placed in such a way that all smaller elements as compared to key are on one side and larger than key are on other side. Then likewise list is broken into two, again algorithm is applied.

2^H+1 - 1, gives total number of nodes present in any binary tree for tree at level 0. 2^H gives the total number of nodes at any level. If root is at level zero, 2^H+1 will give the same, e.g. at level 3, total no. of nodes will be 8 and total number of nodes at level 2 if root is at level 0, is 7. So this relation works here 2^H+1 - 1.

  For static memory location, stack is used. Stack is used because size of memory is known.

While inserting the element make sure the difference of height of  left subtree and right sub tree must be -1, 0 and +1. Insert 10 as a root, then 11 as right child. Here, tree is balanced. Now Insert 12, tree became unbalanced as the height of left sub tree from root and right sub tree is -2, 0 = -2. So do RR rotation on tree. Now 11 is as root and 10 as left child and 12 as right child. Again insert 13 as left child of 12, then 14 right child of 13 again it unbalanced. So do again RR rotation and so on.

Input restricted queue allows input data from one end, and delete data from both ends.

Binary search tree has all the small data in left and all big data in right. So, when inorder traversal is made, it gives sorted list. Suppose tree is having data 1, 2, 3, 4 and 5, if 3 is the root element then 4 and 5 will be on right side of the root. 1 and 2 will get place in left side of 3. So, inorder traversel of BST will give sorted elements.

2^H - 1, gives total number of nodes present in any binary tree. 2^H gives the total number of nodes at any level. So for height H, total number of nodes will be one less than next level. For example at level 3, the number of nodes is 8, for height 3 the total number of nodes is 7. This is the way how this expression works.

Stack is used to perform recursion. It stores the function call in stack. then call each function in LIFO order. Let a function is called from its own body then for to track the function call, each call is pushed into stack till the end of counter then one by one each function call is executed. So, stack is a correct answer.

In stack, the element is inserted in last which should be taken out first. For example, for sequence 4, 3, 5, 2 and 1, first insert 1, 2, 3, 4 then take out 4 and 3; here stack content is 2 and 1, again insert 5 and take out 5, 2 and 1. After insertion of 5 stack content is 5, 2, 1 then take out 5, 2 and 1. At last, we got the sequence 4, 3, 5, 2 and 1.

 For run time memory location, heap is used. As we don't know the size of memory required to allocate. So, heap is used to allocate memory.

2nCn/n+1 where n are the number of nodes.  2nCn/n+1 where n are the number of nodes. For 3 unlabelled nodes it will work as follows ^2*3C₃ / 4= ⁶C_{3 / 4.} So on solving this we get 5, and answer for 5 unlabelled nodes will be 42, so answer 42 is correct.

The data structure which is used to manipulate data is called self referential structure. Graph is a self referential structure. So answer is correct as in Graph structure, we cannot refer all the vertices in some order.

In link list, when value is assigned, its link part is updated. As after step 2, P is pointing to C, after third step, B is pointing to D, after 4^th step, D's link is updated with C and in 5^th step, P is pointing to C, so C will be printed. In linklist when value is assigned, Its link part is updated.

N(H min)=1+N(H-1)+N(H-2). For N(1)=1, for N(2)=2 , for N(3)= 1 + N(2 )+ N(1), it will become 1+2+1 = 4 and so on. So for N(8) it will be 54 and answer 54 is correct.  

Three dimensional array consists of 2D array. First 2 is indicating there are two matrices. So A(0)(0)(1 ) is indicating to 2. A(1)(1)(1) can be written as (((*a +1)+ 1)+1 ), so A(0)(0)(1 ) is referring to second memory location, which is 2. So option as 2 is correct.

In queue, deletion takes place from front end, and rear end is used to insert the element into queue. So this option is correct one.

In stack, insertion and deletion takes place from top.

1. In POST order traversal, first traverse left, then right and then print node. 2. If the sequence to traversal is given then it is easy to predict the tree. In preorder, the node which is traversed first is the root of the tree. So, here A is the root. Now take each node from preorder list and now from list in order the node which is right side of A is right and the node which is on left will come in left. Now traverse one by one each element from the list and make a tree and then find post order traversal from first step. On traversing the tree, we get following sequence for post order: C B E F D A. So the option C B E F D A is correct.

In queue, deletion takes place from front end, and rear end is used to insert the element into queue. So this option is correct.

 It will be printed in ascending order as a root node has all the smaller elements in its left side and all the larger elements in its right side. Likewise it is arranged for each node, always left child is smaller and right child is larger. So, it will arrange all elements in ascending order. So it is correct option.

 If a binary tree contains either zero or two child, then it is called as strict binary tree.

Here 'i' is the stack number, if stack i is full then bottom of new stack begins from there, applicable for multilevel queue. For multilevel queue (n/m)*i - 1, where n is the size of array, m is the size of stack and i is the stack number, let n =9, m=3, then three stacks can be implemented on it. So i = 0, 1, 2. Put i in formula (n/m)*i - 1 for 0 we get -1, -1 is bottom of first stack. Put i=1, we get 2, and 2 is the top for first stack and bottom for second stack. Option says top of first queue is equal to bottom of second queue. So, it is the correct option. As first stack is full, this satisfies the overflow condition.

If (P), then P is pointing to link list, where P is the first pointer. If (!P), then P is pointing to nothing. So, in the while condition, P is satisfying the condition.